Question

If ${lim}_{x\to 0}(\frac{\sin 2x}{x^3}+a+\frac{b}{x^2})=0$ then the value of $3a+b$ is

• A. $2$
• B. $-2$
• C. $-1$
• D. $0$

Answer 1

The correct option is A $2$

Given that ${lim}_{x\to 0}(\frac{\sin 2x}{x^3}+a+\frac{b}{x^2})=0$

This can be written as

${lim}_{x\to 0}\frac{\sin 2x+bx}{x^3}+a=0$

This limit is in $\frac{0}{0}$ form. Applying L'Hospital's rule we get,

$\Rightarrow {lim}_{x\to 0}\frac{2\cos 2x+b}{3x^2}+a=0$

Substituting $x=0$ we get, denominator$=0$. To able able to apply L'Hospital's Rule, Numerator $=0$

$\Rightarrow 2+b=0$

$\Rightarrow b=-2$

$\Rightarrow {lim}_{x\to 0}\frac{2\cos 2x-2}{3x^2}+a=0$

This limit is in $\frac{0}{0}$ form. Applying L'Hospital's rule we get,

$\Rightarrow {lim}_{x\to 0}\frac{-4\sin 2x}{6x}+a=0$

This limit is in $\frac{0}{0}$ form. Applying L'Hospital's rule we get,

$\Rightarrow {lim}_{x\to 0}\frac{-8\cos 2x}{6}+a=0$

$\Rightarrow \frac{-8}{6}+a=0$

$\Rightarrow a=\frac{8}{6}=\frac{4}{3}$

Therefore, $\Rightarrow 3a+b=3\left(\frac{4}{3}\right)-2$

$\Rightarrow 3a+b=4-2=2$